How much discrepency would I get if I used acceleration due to gravity?

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Discussion Overview

The discussion revolves around the potential error in using a standard value of 9.8 m/s² for the acceleration due to gravity in an experiment involving a falling object. Participants explore the implications of this choice in terms of measurement accuracy, the applicability of the equations used, and the effects of air resistance on different types of balls dropped from a height.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the significance of using 9.8 m/s² for a falling object and asks about the potential error in distance calculations using the equation d = -1/2 * a * t².
  • Another participant points out the importance of significant figures in measurements and asks for clarification on the source of error being examined.
  • A participant raises the issue of air resistance, suggesting that while it may be negligible for heavier objects like a cricket ball, it could significantly affect lighter objects such as a ping pong ball.
  • Detailed calculations are provided for both a cricket ball and a ping pong ball, comparing the forces due to gravity and air resistance to illustrate the differing impacts of air resistance on the two objects.

Areas of Agreement / Disagreement

Participants express differing views on the significance of air resistance in the context of the experiment, with some suggesting it is negligible for certain objects while others argue it could be substantial for lighter objects. The discussion remains unresolved regarding the overall impact of using 9.8 m/s² in the calculations.

Contextual Notes

Participants have not reached a consensus on the extent of error introduced by using 9.8 m/s², and the discussion highlights the dependence on factors such as object mass and air resistance, which are not uniformly applicable across different scenarios.

sodium40mg
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How much error would I get if I used 9.8 m/s^2 as the acceleration for a falling object in an experiment? For example, if I have a ball at, say, shoulder's length up in the air and I had someone measure the time it took for the ball to hit the ground. Would I get a lot of error if I used the equation
d = vi * t - 1/2 * a * t^2

Where vi = 0 m/s, so d = -1/2 * a * t^2 and a = - 9.8 m/s^2

Would that actually equal the distance from the shoulder length to the ground or how much error would I get? Would it be significant?
 
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Well exactly where are you trying to examine the erorr from? The fact that you're only using 2 significant figures?
 
sodium40mg said:
How much error would I get if I used 9.8 m/s^2 as the acceleration for a falling object in an experiment? For example, if I have a ball at, say, shoulder's length up in the air and I had someone measure the time it took for the ball to hit the ground. Would I get a lot of error if I used the equation


Would that actually equal the distance from the shoulder length to the ground or how much error would I get? Would it be significant?

Are you talking about measurement error in the timing?

Or are you talking more fundamentally about the applicability of the equation to the real world problem? The biggest discrepancy in the fundamental equation would be that it ignores air resistance. For a ball dropped from about head height the velocity is fairly low so for in most cases the effect of air resistance would be small, but if the ball was extremely low in mass then air resistance could significantly alter the result. In other words, the formula should be quite accurate for a cricket ball but not necessarily so for a ping pong ball.
 
Here's a quick test of whether air resistance will be significant.

Final velocity (ideally) when dropped from 1.8m = 5.9 m/s

Air density at 20C, rho = 1.2 kg/m^2

1. Cricket ball. m = 0.16 kg, frontal area A = 0.0041 m^2, coeff of drag, cd = 0.47.

Air resistance at final velocity, F = 1/2 rho cd A v^2 = 0.041 N

Comparing the above with the force due to gravity of 1.6N we see that the drag is negligible.

2. Pin pong ball. Mass = 0.0027 kg, Frontal area A = 0.00126 m^2, drag coeff cd = 0.47

Air resistance at (ideal) final velocity, F = 1/2 rho cd A v^2 = 0.0125 N.

Comparing the above with the gravitational force of just 0.027 N we see that air resistance would be very significant in this case.
 

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